I want to know how I should visualize modules in algebraic geometry. The way we visualize rings, via their spectra, automatically (or by the beauty of its design) depicts primary decomposition of ideals: the primary components of an ideal $I \triangleleft A$ cut out "primary subschemes" (irreducible and embedded components) whose union is $Z(I)=Spec(A/I)$. (See, for example, Eisenbud and Harris, The Geometry of Schemes, II.3.3, pp. 66-70). This aspect of scheme theory is essential to what makes it "geometric."

By this standard, I think however we visualize modules should allow us to depict primary decomposition of submodules; otherwise I would say it's not a very good visualization.

If we're happy taking quotients, WLOG we can just look at primary decompositions of $0$. So let $M$ be a finitely generated module over a Noetherian ring $A$, and $0=N_1\cap\cdots\cap N_n$ be a primary decomposition of $0$ in $M$, with primes $P_i$ co-associated to the primary modules $N_i$, i.e. associated to the coprimary modules $M/N_i$.

How can one visualize the modules $M,N_1,\ldots,N_n$ in relation to $Spec(A)$ in a way that meaningfully depicts:
(1) the primary decomposition of $0$ in $M$ (in particular that the $N_i$ are primary in $M$), and
(2) the relationship of the modules $N_i$ to their co-associated primes, say
{ $P_i$ } $ = Ass(M/N_i) \subseteq Spec(A)$?

Some useful background results to make sense of the above (all rings and modules are Noetherian):

  • The primes $P_i$ co-associated to $N_i$ are precisely the associated primes of $M$ (see R. Ash, Comutative Algebra, Theorem 1.3.9)

  • A module $Q$ is coprimary iff it has exactly one associated prime $P$, and then $P=\sqrt{ann Q}$. (see R. Ash, Comutative Algebra, Corollary 1.3.11)


I will assume that everything in sight is Noetherian and finitely generated. Then the primary decomposition amounts to a description of the geometric support of the module $M$ if you view it as a sheaf on $\text{Spec}(A)$. The scheme $\text{Spec}(A/\text{Ann}(M))$ is a subscheme of $\text{Spec}(A)$ that, by definition, supports $M$. If $N_i$ is minimal, then $\text{Spec}(A/P_i)$ is an irreducible component of $\text{Spec}(A/\text{Ann}(M))$. The module $M/N_i$ is also a quotient of $M \otimes (A/P_i^n)$ for $n$ large enough (and in a reduced decomposition I think they are equal), so it is part of $M$ on that irreducible component of its support. If $N_i$ is not minimal, then $\text{Spec}(A/P_i)$ is an irreducible scheme inside of a component of $\text{Spec}(A/\text{Ann}(M))$.

  • $\begingroup$ Seems like it's not mentioned that any irreducible component arises as $\Spec(A/P_i)$, so I'm adding this remark. Another thing: While minimal primes are in general easier to visualize, it seems really hard to depict embedded primes. Does anyone have good ideas on that? $\endgroup$ Dec 7 '09 at 13:38
  • 2
    $\begingroup$ The visualization is easy: The embedded primes are dots on top of the curves drawn in the surface, or curves on top of the surfaces drawn in space. $Spec(A/P_i)$ is a subvariety embedded in the subvariety $Spec(A/P_j)$. $\endgroup$ Dec 7 '09 at 14:59
  • $\begingroup$ @Ho Chung, definitely check out the references I made to The Geometry of Schemes. @Greg, thanks, as usual! In case you're wondering what's up, I want to leave the question open awhile longer until I've had more time to think about it myself, in case someone might mentions something else cool I should know. I haven't forgotten about it :) Cheers, $\endgroup$ Dec 10 '09 at 8:19

Visualizing embedded primes: In P^2, a one dimensional scheme cannot have embedded points unless its ideal has more than one generator, by the unmixedness theorem of Macaulay. So imagine we have two polynomials that define a one dimensional scheme in P^2. We will imagine this scheme as a limit of zero dimensional schemes. First take two quadratic polynomials, one of which is a product of two linear factors, i.e. take one pair of lines meeting at p, and another irreducible conic. In general the irreducible conic C meets each of the lines twice, away from p. Thus the two qudratic polynomials define a zero dimensional scheme of 4 points.

Now hold fixed the two intersections of C with one of the lines L, and let the two intersections of C with the other line M approach p, i.e. let C become tangent to M at p. When this occurs, the conic C now contains three distinct points of L, hence C has become reducible and contains L. Now the scheme defined by intersecting L+M with C has become one dimensional, reducible, and consists set theoretically only of the line L. I claim the point p is an embedded point of the component L of the scheme defined by L+M and C.

This is easy algebraically, since the ideal of the given scheme is (xy,(x(x-y)) = (x^2, xy) which is the intersection of the primary ideals (x) and (x^2, xy, y^2), with associated primes (x) and (x,y). Hence (x,y) is an embedded prime. I.e. the origin is an embedded point on the y axis for this scheme. This also helps explain the apparent failure of Bezout's theorem for this intersection of two conics apparently not having degree 4.

In general, in P^n, a scheme S with embedded subschemes must be defined by intersecting more hypersurfaces than the codimension of S. Thus such an S can always be viewed as a limit of lower dimensional schemes. It seems to me that embedded subschemes should arise when these lower dimensional schemes are reducible and some lower dimensional component comes to lie on a larger dimensional component of the limit. I do not know if this intuition is the only possibility, and since the world is wide, probably not.

  • $\begingroup$ I wonder where this limit process is taking place. Certainly not in $Hilb \mathbb{P}^n$, because of the jump in dimension. Is it just a special fiber in a non-flat family of homogeneous ideals in k[x,y,z]? $\endgroup$
    – quim
    Nov 17 '10 at 11:00
  • 2
    $\begingroup$ We are brainwashed into thinking the only families in the world are flat. Given any morphism, each special fiber is, in a sense defined by that morphism, a limit of nearby fibers, whether they have the same dimension or not. It just isn't the limit from one direction. Part of it comes from each direction. E.g. when you blow up a point, the exceptional fiber is a limit of all nearby fibers, each point of the exeptional fiber arising from limits in a different direction. To see this from a flat perspective, blow up the map so it becomes flat, so the big fiber becomes a union of fibers . $\endgroup$
    – roy smith
    Nov 17 '10 at 23:36
  • $\begingroup$ Thus the parameter space seems to be a blow down of a hilbert scheme. Or in the present example, you could just use the product HxH of two copies of the space H (hilbert scheme) of plane conics. For a universal family you just use the corresponding incidence sub variety of HxHxP^2. I.e. triples <f,g,p> such that f(p) = g(p) = 0. $\endgroup$
    – roy smith
    Nov 17 '10 at 23:46
  • $\begingroup$ oh, i'm sorry, now i see your last sentence. i would say yes. $\endgroup$
    – roy smith
    Nov 18 '10 at 1:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.